Question

$ \underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{-1}}x-x}{{{x}^{3}}\,\cos x} $ is equal to

• A. $ 1/2 $
• B. $ 1/3 $
• C. $ 1/6 $
• D. $ 1/12 $

Answer 1

Answer: (C)

$ =\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{-1}}x-x}{{{x}^{3}}\cos x} $
$ \left( \frac{0}{0}\,form \right) $
$ =\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{1}{\sqrt{1-{{x}^{2}}}}-1}{{{x}^{3}}(-\sin x)+3{{x}^{2}}\cos x} $
(using L- Hospital's rule)
$ =\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\sqrt{1-{{x}^{2}}}}{\sqrt{1-{{x}^{2}}}.{{x}^{2}}(-x\,\sin \,x+3\,\cos \,x)} $
$ \times \frac{1+\sqrt{1-{{x}^{2}}}}{1+\sqrt{1-{{x}^{2}}}} $
$ =\underset{x\to 0}{\mathop{\lim }}\,\frac{1-1+{{x}^{2}}}{\left[ \begin{align} & {{x}^{2}}\sqrt{1-{{x}^{2}}}(1+\sqrt{1-{{x}^{2}}}) \\ & (-x\,\sin \,x+3\,\cos \,x) \\ \end{align} \right]} $
$ =\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\left[ \begin{align} & \sqrt{1-{{x}^{2}}}(1+\sqrt{1-{{x}^{2}}}) \\ & (-x\,\sin \,x+3\,\cos \,x) \\ \end{align} \right]} $
$ =\frac{1}{1(1+1)\,(3)} $
$ =\frac{1}{6} $